Find the gradient and the y-intercept of each of the following lines:

(i) 5x – 10y = 3

(ii) $\dfrac{x}{6} + \dfrac{y}{9} = 1$

(iii) x + 4 = 0

(iv) y = 6

(i) Given,

⇒ 5x – 10y = 3

Converting 5x – 10y = 3 in the form y = mx + c we get,

⇒ -10y = 3 - 5x

⇒ y = $-\dfrac{3}{10} - \dfrac{5x}{-10}$

⇒ y = $-\dfrac{3}{10} + \dfrac{x}{2}$

⇒ y = $\dfrac{1}{2}x - \dfrac{3}{10}$

The equation of straight line is given by, y = mx + c, where m is the slope and c is the y-intercept.

Comparing, y = mx + c with y = $\dfrac{1}{2}x - \dfrac{3}{10}$, we get:

m = slope = $\dfrac{1}{2}$

c = y-intercept = $-\dfrac{3}{10}$

Hence, slope = $\dfrac{1}{2}, \text{y-intercept} = -\dfrac{3}{10}$.

(ii) Converting $\dfrac{x}{6} + \dfrac{y}{9} = 1$ in the form y = mx + c we get,

$\Rightarrow \dfrac{x}{6} + \dfrac{y}{9} = 1 \\[1em] \Rightarrow \dfrac{y}{9} = 1 - \dfrac{x}{6} \\[1em] \Rightarrow y = 9 \Big(1 - \dfrac{x}{6}\Big) \\[1em] \Rightarrow y = \Big(9 - \dfrac{9x}{6}\Big) \\[1em] \Rightarrow y = -\dfrac{3x}{2} + 9.$

The equation of straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

Comparing y = mx + c with $y = -\dfrac{3x}{2} + 9$, we get:

m = slope = $-\dfrac{3}{2}$

c = y-intercept = 9

Hence, slope = $-\dfrac{3}{2}$, y-intercept = 9.

(iii) Given,

x + 4 = 0

x = -4

This is a vertical line parallel to the y-axis, passing through the x-axis at x = -4.

We know that the inclination of a line parallel to y-axis is 90°.

∴ Slope of y-axis = tan 90° = infinity, which is not defined.

Since the line is parallel to the y-axis and passes through a negative x-value, it never crosses the y-axis. There is no y-intercept.

Hence, slope is not defined and line has no y-intercept.

(iv) Given,

y = 6

This is a horizontal line parallel to the x-axis, passing through the y-axis at y = 6.

We know that the inclination of a line parallel to x-axis is 0°.

∴ Slope of a line parallel to x-axis = tan 0° = 0.

y-intercept = 6

Hence, slope = 0 and y-intercept = 6.